Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $y = \dfrac{9q + 27}{q + 10} \times \dfrac{-q - 10}{q^2 + q - 6} $
Answer: First factor the quadratic. $y = \dfrac{9q + 27}{q + 10} \times \dfrac{-q - 10}{(q + 3)(q - 2)} $ Then factor out any other terms. $y = \dfrac{9(q + 3)}{q + 10} \times \dfrac{-(q + 10)}{(q + 3)(q - 2)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ 9(q + 3) \times -(q + 10) } { (q + 10) \times (q + 3)(q - 2) } $ $y = \dfrac{ -9(q + 3)(q + 10)}{ (q + 10)(q + 3)(q - 2)} $ Notice that $(q + 10)$ and $(q + 3)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -9\cancel{(q + 3)}(q + 10)}{ (q + 10)\cancel{(q + 3)}(q - 2)} $ We are dividing by $q + 3$ , so $q + 3 \neq 0$ Therefore, $q \neq -3$ $y = \dfrac{ -9\cancel{(q + 3)}\cancel{(q + 10)}}{ \cancel{(q + 10)}\cancel{(q + 3)}(q - 2)} $ We are dividing by $q + 10$ , so $q + 10 \neq 0$ Therefore, $q \neq -10$ $y = \dfrac{-9}{q - 2} ; \space q \neq -3 ; \space q \neq -10 $